Question

Solve using a formula. Please don't guess, I would like professionals to take a look

Answer 1

Let's take this problem step by step:

What we know:

 $x+y=4\\xy=-2$

Before we solve, let's do one thing that will help us out greatly later down the road:

 $x+y=4\\(x+y)^2=4^2\\x^2+2xy+y^2=16\\x^2+2(xy)+y^2=16\\x^2+2(-2)+y^2=16\\x^2+4+y^2=16\\x^2+y^2=20$<--- useful equation

Let's rearrange the problem a little bit:

 $x+\frac{x^3}{y^2}+\frac{y^3}{x^2} +y=\frac{x^3}{x^2} +\frac{x^3}{y^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}$

Combine fractions of common denominators:

$\frac{x^3+y^3}{x^2} +\frac{x^3+y^3}{y^2} =(x^3+y^3)*(\frac{1}{x^2}+\frac{1}{y^2} )$

Now's let factor everything apart:

 $(x^3+y^3)=(x+y)(x^2-xy+y^2)\\\\\frac{1}{x^2}+\frac{1}{y^2} =\frac{x^2+y^2}{x^2y^2}$

Let's use what we know and our useful equation:

 $(x+y)*(x^2-xy+y^2)*(\frac{x^2+y^2}{x^2y^2} )\\=4*(x^2+y^2-xy)*(\frac{20}{(xy)^2} )\\=4*(20-(-2))*\frac{20}{(-2)^2} \\=4*22*5\\=440$

The value is 440

Answer: 440

Hope that helps!

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